Question

A vessel contains dimethyl ether at a pressure of $0.4atm$. Dimethyl ether decomposes as $C{H}_{3}OC{H}_{3\left(g\right)}\to C{H}_{4\left(g\right)}+C{O}_{\left(g\right)}+H_{2\left(g\right)}$. The rate constant of decomposition is $4.78\times {10}^{-3}mi{n}^{-1}$. Calculate the ratio of initial rate of diffusion to rate of diffusion after $4.5$ hours of initiation of decomposition. Assume the composition of gas present and gas diffused to be same.

Answer 1

The pressure of dimethyl ether after 4.5 hours will be $A=A_0{e}^{-kt}=0.4\times e^{-4.78\times {10}^{-3}\times 4.5\times 60}=0.11$
Total pressure after 4.5 hours will be $0.4+2(0.4-0.11)=0.98mm$.
Average molecular weight after 4.5 hours will be $\frac{0.4\times 46+0.29\times 16+0.29\times 28+0.29\times 2}{0.98}=18.86$
The ratio of initial rate of diffusion to rate of diffusion after 4.5 hours of initiation of decomposition will be $\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}\times \frac{P_1}{P_2}=\sqrt{\frac{18.86}{46}}\times \frac{0.4}{0.98}=0.261$